Homework Help - Post Questions, Assignments & Papers

Studypool values your privacy. Only questions posted as Public are visible on our website.

AC Assgnment 8

Anonymous

labelOther

timer
Asked: Nov 16th, 2013

Question description

Here is what my instructor said to fix

Resubmit the assignment. Clearly indicate the final answers for all the
work, such as: 16 a) the answer, b) another answer, c)
etc.

23-2R1=12 ΩR2=18 ΩVt=15VRt=R1+R2Rt=12+18=30 ΩI=Vt/Rt=15/30=0.5AV1=IR1=0.5*12=6VV2=IR2=0.5*18=9V(a)Vt and I are in phase(b)V1 and I are in phase(c)V2 and I are in phase

23-4:R1=12 ΩR2=18 ΩVa=36VRt=R1+R212+18=30 ΩI1=Va/R1=36/12=3AI2=Va/R2=36/18=2AIt=I1+I2=3+2=5A(a)Va and I1 are in same phase(b)Va and I2 are In same phase(c)Va and It are in same phase

23-6:Vt=120VXL1=100 ΩXL2=150 ΩXLT=XL1+XL2100+150=250 ΩI=Vt/Xlt=120/250=0.48AV1=IXlt=0.48*100=48V2=IXlt=0.48*150=72The ohms of XL are just as effective as ohms of R in limiting the current or producing a voltage drop.XL has a phasor quantity with a 90° phase angle.

23-8:Va=120VXl1=100 ΩXl2=400 ΩI1=Va/Xl1=120/100=1.2AI2=Va/Xl2=120/400=0.3AIt=I1+I2=0.3+1.2=1.5A(a)Va and I1 are(b)Va and I2 are(c)Va and It areThe ohms of XL are just as effective as ohms of R in limiting the current or producing a voltage drop.XL has a phasor quantity with a 90° phase angle.

23-10Vt=18VXc1=220 ΩXc2=680 ΩV1=IXc1=4.4VV2=IXc2=13.6VXc=220+680I=Vt/Xc=18/900=0.02ASince there is no R or XL, the series ohms of XC can be combined directly.

23-12:Vt=10VXc1=100 ΩXc2=25 ΩI1=Va/Xc1=10/100=0.1AI2=Va/Xc2=10/50=0.2AIt=0.2+0.1=0.3ASince there is no R or XL, parallel IC currents can be added.